- Thread starter
- #1

and [tex]m =[/tex] no. of terms in the expression [tex]y[/tex]

and [tex]n =[/tex] no. of integers for which [tex]y[/tex] has min. value

Then [tex]\displaystyle\frac{m+n-18}{10} =[/tex]

- Thread starter jacks
- Start date

- Thread starter
- #1

and [tex]m =[/tex] no. of terms in the expression [tex]y[/tex]

and [tex]n =[/tex] no. of integers for which [tex]y[/tex] has min. value

Then [tex]\displaystyle\frac{m+n-18}{10} =[/tex]

- Jan 26, 2012

- 890

Maybe it's me, but I find that incomprehensible.

and [tex]m =[/tex] no. of terms in the expression [tex]y[/tex]

and [tex]n =[/tex] no. of integers for which [tex]y[/tex] has min. value

Then [tex]\displaystyle\frac{m+n-18}{10} =[/tex]

For a start why is \(m\) not \(5152\)?

Do you mean \(n\) to be the number of integers corresponding to a

(the slope is -5152 for -ve \(x\), and increases by 2 when we pass a integer argument moving to the right ...)

CB

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